Discussion Forum



1)

A random variable X has the probability distribution

    X 

     1

     2  

     3

     4

     5

     6

     7

     8

  p(X)

  0.15

  0.23

  0.12

  0.10

  0.20

  0.08

  0.07

  0.05

For the events E = {X is a prime number} and F={X<4}, then p(E ∪ F) is


A) 0.50

B) 0.77

C) 0.35

D) 0.87

Answer:

Option B

Explanation:

P(E) = P(2 or 3 or 5 or 7)
= 0.23 + 0.12 + 0.20 + 0.07 = 0.62
P(F)= P(1 or 2 or 3)
=0.15 +0.23+ 0.12 = 0.50
P(E ∩ F) = P(2 or 3)
  =0.23+0.12=0.35
.'. P(EUF) = P(E)+ P(F)- P(E ∩ F)
=0.62+ 0.50-0.35 = 0.77